Question

我正在研究MVC应用程序。我试图通过JQuery将参数从视图传递到控制器使用JSON。返回类型工作正常但是当我传递参数时它失败了。以下是代码

查看

<script type="text/javascript">
    $(document).ready(function () {
        $("#ClientList").change(function () {
            $.ajax({
                url: '@Url.Action("GetClientData", "Home")',
                type: 'Post',
                dataType: 'json',
                data: { clientID: $("#CientList").val() },
                contentType: 'application/json; character=utf-8',
                success: function (data) {
                    if (data.success) {
                        document.getElementById("ClientName").value = data.clientName
                    }
                    else {
                        alert('invalid ID' + data.success);
                    }
                }
            });
        });
    });

CONTROLLER

[HttpPost]
        public JsonResult GetClientData(string clientID)
        {
            Client client = Client.GetClient(Convert.ToInt32(clientID));
            if (client != null)
            {
                return Json(new { success = true, clientName = client.ClientName });
            }
            return Json(new { success = false });

        }

如果我从控制器代码中取出参数。它工作正常但是当方法接受参数时失败。有什么建议吗？

Answer 1

<script type="text/javascript">
    $(document).ready(function () {
        $("#ClientList").change(function () {
            var selectedClientID = { clientID: $("#ClientList").val() };
            $.ajax({
                url: '@Url.Action("GetClientData", "Home")',
                type: 'Post',
                contentType: 'application/json',
                dataType: 'json',
                data: JSON.stringify(selectedClientID),
                success: function (data) {
                    if (data.success) {
                        document.getElementById("ClientName").value = data.clientName;
                        document.getElementById("ClientAddress").value = data.clientAddress;
                        document.getElementById("ClientContactNumber").value = data.clientContact;
                    }
                    else {
                        alert('invalid ID' + data.success);
                    }
                }
            });
        });
    });
</script>

这完美无缺

将参数从视图传递给控制器

1 个答案: